Why does damping affect the time period of SHM oscillators?

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SUMMARY

The discussion centers on the impact of damping on the time period of simple harmonic motion (SHM) oscillators, specifically springs and pendulums. The equations for the time period of a spring (T = 2π * √(m/k)) and a pendulum (T = 2π * √(l/g)) indicate that the time period is independent of amplitude; however, damping significantly alters this relationship. Damped systems do not undergo true SHM, leading to discrepancies between theoretical equations and real-world behavior. The time period in a damped system can change over time, particularly as damping increases.

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  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with the equations governing springs and pendulums
  • Knowledge of damping mechanisms in oscillatory systems
  • Basic grasp of Fourier transforms for analyzing motion equations
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jsmith613
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There are two equations that can describe the time period of SHM oscillators (springs / pendulums ONLY)

Spring T = 2π * \sqrt \frac {m}{k}

Pendulum T = 2π * \sqrt \frac {l}{g}

It would seem from these equations that time period is independent of amplitude
therefore we should be able to conclude that the time period for a damped oscillator is the same as that for an undamped oscillator.

BUT if you oscillate something viscous (water, oil, tar) then then time period is massively increased.

My questions are
(a) would an underdamped system oscillate at the same period as an un-damped system?
(b) how come the equations of SHM and reality don't match up (damped systems are still undergoing SHM...I think)
(c) will the time period change in the damped system during the oscillations? i.e: could it start at 0.5s and then increas to 1s
 
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First point - the equations you state are approximations. The spring one assumes the spring is within its elastic limit and Hookes law strictly applies (not to bad a condition in the real world). The pendulum equation is more of a problem. It is only valid if the pendulum oscillation is through an infinitestimaly small angle.
So real a real pendulum does not execute true SHM and its period is affected by its amplitude.

In answer to your questions

a) No and the period changes with time
b) Damped systems are not undergoing true SHM hence the mismatch
c) Yes

If you are familiar with Fourier transforms you can take the transform of the equation of motion of the damped system and see the change in period directly.

Hope this helps

Regards

Sam
 
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The way the system behaves depends how it is damped. Different physical processes can produce damping forces that are constant amplitude (but change direction every half cycle of oscillation), proportional to velocity, or proportional to veloicty squared - or even more complicated behaviour.

The simplest to analyse is damping proportioal to velocity, and in that case the damped frequency is constant for the whole motion, but is lower than the frequency without damping. For systems with low levels of damping like a mass oscillating on a spring, the change in frequency is small (typically less than 1%) but as the damping increases the "frequency" eventually drops to zero and there is no oscillation at all, just a motion in one direction that slows down as the system approaches its equilibrium position.

All this is well known and covered in college-level courrses on dynamics, but as in any subject you have to learn to walk before you can run, and a first course in dynamics often only includes undamped SHM.
 
thanks all
 

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